Computing this Lie derivative more explicitly [1]}, we have that
\(£_{\bar{\mathbf {V}}}\,\mathbf {\Omega } = \left(\tilde{\mathbf {d}}\mathbf {\Omega }\right)\left(\bar{\mathbf {V}}\right)+\tilde{\mathbf {d}}\left(\mathbf {\Omega }\left(\bar{\mathbf {V}}\right)\right) = \tilde{\mathbf {d}}\left(\mathbf {\Omega }\left(\bar{\mathbf {V}}\right)\right)=0,\) 
